输入特征以检查类是否具有成员函数 [英] Type traits to check if class has member function
问题描述
尝试创建一种方法来识别给定的类是否具有可以调用的给定函数,并返回某种类型。
Trying to create a way to identify if a given class has a given function that can be invoked, and returns some type.
对我所要表达的任何想法这里做错了吗?
Any idea on what I'm doing wrong here? Is there a better way to determine if a given method is invoke'able given a class?
#include <string>
#include <type_traits>
#define GENERATE_HAS_MEMBER_FUNC(func, rettype) \
template<typename T, class Enable = void> struct has_##func; \
template<typename T, class U> struct has_##func : std::false_type {}; \
template<typename T> \
struct has_##func<T, \
typename std::enable_if<std::is_same< \
typename std::result_of<decltype (&T::func)(T)>::type, \
rettype>::value>::type> : std::true_type{}; \
template<class T> constexpr bool has_##func##_v = has_##func<T>::value;
GENERATE_HAS_MEMBER_FUNC(str, std::string)
GENERATE_HAS_MEMBER_FUNC(str2, std::string)
GENERATE_HAS_MEMBER_FUNC(funca, std::string)
GENERATE_HAS_MEMBER_FUNC(strK, std::string)
GENERATE_HAS_MEMBER_FUNC(fancy, std::string)
GENERATE_HAS_MEMBER_FUNC(really, std::string)
struct A1 {
virtual std::string str() const { return ""; }
std::string strK() const { return ""; }
virtual std::string fancy()=0;
};
struct A2 : A1 {
std::string str() const override { return ""; }
std::string funca();
std::string fancy() override { return ""; }
std::string really(int a=0) const { return std::to_string(a); }
};
int main() {
static_assert(has_str_v<A1>,
"A1::str is virtual method with impl on base"); // MSVC: NO, clang: OK, GCC: NO
static_assert(has_strK_v<A1>,
"A1::strK is implemented inline "); // MSVC: NO, clang: OK, GCC: NO
static_assert(has_fancy_v<A1>,
"A1::fancy is a pure virtual method on base"); // MSVC: NO, clang: OK, GCC: NO
static_assert(!has_really_v<A1>,
"A1::really doesn't exist in A1"); // MSVC: OK, clang: OK, GCC: OK
static_assert(has_str_v<A2>,
"A2::str is override method "); // MSVC: OK, clang: OK, GCC: OK
static_assert(!has_str2_v<A2>,
"A2::str2 does not exist in A2"); // MSVC: NO, clang: OK, GCC: OK
static_assert(has_funca_v<A2>,
"A2::funca is defined (no impl) in A2"); // MSVC: OK, clang: OK, GCC: OK
static_assert(has_strK_v<A2>,
"A2::strK is implemented method on base"); // MSVC: OK, clang: OK, GCC: OK
static_assert(has_fancy_v<A2>,
"A1::fancy is a override of pure virtual method of base"); // MSVC: OK, clang: OK, GCC: OK
static_assert(has_really_v<A2>,
"A2::really has default param (can be invoked without params)"); // MSVC: OK, clang: NO, GCC: NO
return 0;
}
此实现有些令人惊讶。
Some surprises on this implementation.
编辑:
在尝试实现@ Jarod42和@Vittorio Romeo时,提出了很棒的建议:
Upon trying to implement @Jarod42 and @Vittorio Romeo awesome suggestions:
#define GENERATE_HAS_MEMBER_FUNC(func, rettype) \
template<class T> using _has_##func_chk = \
decltype(std::declval<T &>().func()); \
template<class T> constexpr bool has_##func##_v = \
is_detected_exact_v<rettype, _has_##func_chk, T>;
现在VS2015上有两个测试用例仍然失败(没有任何意义):
static_assert(!has_really_v, A1 :: really在A1中不存在);
static_assert(!has_str2_v, A2 :: str2在A2中不存在);
now two test cases still fail on VS2015 (doesn't make any sense): static_assert(!has_really_v, "A1::really doesn't exist in A1"); static_assert(!has_str2_v, "A2::str2 does not exist in A2");
我可能缺少一些愚蠢的东西。
there is probably something silly something that I'm missing... any clues?
推荐答案
有没有更好的方法来确定给定方法是否被调用?能给定一堂课吗?
Is there a better way to determine if a given method is invoke'able given a class?
是的,您可以使用 检测习惯用法 ,可以在C ++ 11 中实现(链接页面包含有效的实现)。
下面是一个例子: Cat
是否有 float Cat :: purr (int)
方法?
Here's an example: does Cat
have a float Cat::purr(int)
method?
struct Cat { float purr(int){} };
template<class T>
using has_purr =
decltype(std::declval<T&>().purr(std::declval<int>()));
static_assert(std::experimental::is_detected_exact_v<float, has_purr, Cat>);
必需的检测习惯用法 C ++ 17依赖关系在C ++ 11中很容易实现:
The required detection idiom C++17 dependencies are trivial to implement in C++11:
template< class... >
using void_t = void;
struct nonesuch {
nonesuch() = delete;
~nonesuch() = delete;
nonesuch(nonesuch const&) = delete;
void operator=(nonesuch const&) = delete;
};
这是 魔盒上完全符合C ++ 11的最小示例 。
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